# Expectation of time integral

I've 2 questions:

let be $W_s$ a standard Brownian motion:

• using Ito's formula show that $\left( W_t,\int_0^t W_sds \right)$ has a normal distribution;
• and calculate $E\left[e^{W_t}e^{\int_0^t W_sds} \right] .$

For the first part, i know that $W_t$ and $\int_0^t W_sds$ have normal distribution with mean and variance respectively $(0,t)$ and $(0, t^3/3)$, but i need help with Ito's formula.

For the second part i've tried to solve $E\left[e^{W_t}e^{\int_0^t W_sds} \right]= \iint e^{W_t}e^{\int_0^t W_sds} \;\phi \left( W_t,\int_0^t W_sds \right)\: dW_t \int_0^t W_sds$...

Is these the only way?

P.S. sorry for my poor english

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What does $(W_t , \int_0^t W_s ds)$ mean? – Raskolnikov Jan 10 '11 at 10:48
@Raskolnikov, probably vector with first element $W_t$ and second $\int_0^tW_sds$. OP problem probably is to show that this vector is bivariate normal. – mpiktas Jan 10 '11 at 11:29

Let me expand on mpiktas answer. As you noted, $W_t$ and $\int_0^t {W_s ds}$ are zero mean normal with variance $t$ and $t^3/3$, respectively (the first is trivial). Now, if $(W_t,\int_0^t {W_s ds})$ is bivariate normal, then any linear combination $aW_t + b\int_0^t {W_s ds}$, $a,b \in \mathbb{R}$, is univariate normal. In particular, the sum $W_t + \int_0^t {W_s ds}$ is (zero mean) normal. Since you are given the variance of $W_t$ and $\int_0^t {W_s ds}$, the variance of the sum follows straight from the covariance ${\rm Cov}(W_t,\int_0^t {W_s ds})$, that is from ${\rm E}[W_t \int_0^t {W_s ds}]$. Now, assuming you found that variance, calculating ${\rm E}[\exp (W_t + \int_0^t {W_s ds} )]$ amounts to calculating the moment-generating function at $1$ of a ${\rm N}(0,\sigma^2)$ variable.

EDIT: Regarding the first question, let's explain why we should expect $(W_t,\int_0^t {W_s ds})$ to be (bivariate) normal. Since $W$ is continuous on the interval $[0,t]$, $\int_0^t {W_s ds}$ is a standard Riemann integral, and $\int_0^t {W_s ds} \approx \frac{t}{n}\sum\nolimits_{k = 0}^{n - 1} {W_{kt/n} }$. Denote the right-hand side by $Y_t^n$. Then $(W_t,Y_t^n)$ is normal. Indeed, $W$ is a Gaussian process; hence, by definition, $(W_{t_1},\ldots,W_{t_m})$ is normal for any choice of times $t_1,\ldots,t_m$. But $(W_{t_1},\ldots,W_{t_m})$ is normal if and only if any linear combination $\sum\nolimits_{i = 1}^m {a_i W_{t_i } }$, $a_i \in \mathbb{R}$, is (univariate) normal, and the conclusion follows. Finally, since $(W_t,Y_t^n)$ is normal, we should expect that in the limit as $n \to \infty$, $(W_t,\int_0^t {W_s ds})$ is normal.

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If you have already shown that $W_t$ and $\int_0^tW_sds$ are normal, find their covariance and use moment generating function of bivariate normal distribution to solve the second part of your problem.

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If i've understood what u say, i've to calculate $E[W_t \int_0^t W_s ds]$.

Tell me if i'm right doing this:

as $\int_0^t W_sds=tW_t-\int_0^t s dW_s$, i've to calculate $E[W_t (tW_t-\int_0^t s dW_s)]$, or better $E[t W^2_t]-E[W_t\int_0^t s dW_s)]$, and the first expectation is $t^2$.

as $W_t=\int_0^t dW_s$ the second expectation become $E[\int_0^t dW_s\int_0^t s dW_s)]$ that is for Ito-isometry (hope that this is the correct translation) = $E[\int_0^t s ds)]$, and this expectation is $t^2/2$.

So covariance is $t-t^2/2=t^2/2$...

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You could find the covariance directly by $$E[W_t\int_0^t W_s ds]=\int_0^t E[W_t W_s] ds=\int_0^t (t\wedge s) ds=\int_0^t s ds=t^2/2.$$ – Byron Schmuland Jan 11 '11 at 2:11
Your answer is fine, but Byron's approach is much more elementary. – Shai Covo Jan 11 '11 at 4:01
Just note the typo at the end (you meant to write $t^2-t^2/2=t^2/2$). – Shai Covo Jan 11 '11 at 17:22